This is more of a too-long comment than an answer.

The specific question in the comments, "can I find for every finite field, say of characteristic neq 2, an elliptic curve $y^2 = D(x)$, with $D$ of degree 4 and irreducible such that the number of points is even" has a nice positive answer: yes, for *any* irreducible $D$. And the same trick applies to produce hyperelliptic examples in every odd genus.

Let $g \ge 1$ be odd, and let $F$ be an irreducible polynomial of degree $2g+2$ over $\mathbb{F}_q$ with $q$ odd (I will use $F$, rather than $D$, for the polynomial, so that I can use $D$ for divisors). We define a genus-$g$ hyperelliptic curve
$
H: y^2 = F(x)
$
over $\mathbb{F}_q$. Let $D_\infty$ be the (degree 2) divisor at infinity on $H$.

The polynomial $F$ factors over $\mathbb{F}_{q^2}$ as $F = GG'$ where $G$ and $G'$ are irreducible polynomials of degree $g+1$ over $\mathbb{F}_{q^2}$, conjugate over $\mathbb{F}_q$. Let $\{\alpha_1,\ldots,\alpha_{g+1}\}$ be the roots of $G$ and $\{\alpha_1',\ldots,\alpha_{g+1}'\}$ be the roots of $G'$. Since $g$ is odd, we can define divisors $D := \sum_{i=1}^{g+1}(\alpha_i,0) - ((g+1)/2)D_\infty$ and $D' := \sum_{i=1}^{g+1}(\alpha_i',0) - ((g+1)/2)D_\infty$ on $H$, both of degree $0$ and both defined over $\mathbb{F}_{q^2}$. Now $D$ and $D'$ are not principal (they correspond to the ideals $(G(x),y)$ and $G'(x),y)$, but by construction $2D = (G)$ and $2D' = (G')$, so $[D]$ and $[D']$ are 2-torsion divisor classes exchanged by Galois. But in fact they are the same class, because $D + D' = (y)$; so $[D] = [D']$ is an $\mathbb{F}_q$-rational divisor class of order 2, and therefore the divisor class group of $H$ has even order.